(i) To find \(f^{-1}(x)\), start with \(y = 2x + 5\). Solve for \(x\):

\(y - 5 = 2x\)

\(x = \frac{1}{2}(y - 5)\)

Thus, \(f^{-1}(x) = \frac{1}{2}(x - 5)\).

For \(g^{-1}(x)\), start with \(y = \frac{8}{x-3}\). Solve for \(x\):

\(y(x - 3) = 8\)

\(yx - 3y = 8\)

\(yx = 8 + 3y\)

\(x = \frac{8}{y} + 3\)

Thus, \(g^{-1}(x) = \frac{8}{x} + 3\), and it is not defined for \(x = 0\).

(ii) Given \(fg(x) = 5 - kx\), we have:

\(fg(x) = f\left(\frac{8}{x-3}\right) = 2\left(\frac{8}{x-3}\right) + 5 = \frac{16}{x-3} + 5\)

Set \(\frac{16}{x-3} + 5 = 5 - kx\)

\(\frac{16}{x-3} = -kx\)

\(16 = -kx(x-3)\)

\(kx^2 - 3kx + 16 = 0\)

For no solutions, the discriminant must be less than zero:

\(b^2 - 4ac = (-3k)^2 - 4(k)(16) < 0\)

\(9k^2 - 64k < 0\)

\(k(9k - 64) < 0\)

The critical points are \(k = 0\) and \(k = \frac{64}{9}\).

Thus, the set of values is \(0 < k < \frac{64}{9}\).

(i) The function f(x) = 4x − 2x² is a quadratic function. The vertex form of a quadratic function ax² + bx + c is given by completing the square or using calculus to find the turning point. The turning point occurs at x = 1, and the maximum value of f(x) is 2. Therefore, the range of f is ≤ 2.

(ii) To find the value of k for which gf(x) = k has equal roots, substitute f(x) into g(x):

gf(x) = g(f(x)) = 5(4x − 2x²) + 3 = 20x − 10x² + 3.

Set this equal to k:

\(−10x² + 20x + 3 = k.\)

For equal roots, the discriminant must be zero:

\(b² − 4ac = 0, where a = −10, b = 20, c = 3 − k.\)

\(20² − 4(−10)(3 − k) = 0.\)

\(400 + 40(3 − k) = 0.\)

\(400 + 120 − 40k = 0.\)

\(520 = 40k.\)

\(k = 13.\)

To find the value of \(k\) for which the equation \(gf(x) = 0\) has two equal roots, we need to substitute \(f(x)\) into \(g(x)\).

\(gf(x) = g(f(x)) = 2(2x^2 - 12x + 7) + k = 0\)

This simplifies to:

\(4x^2 - 24x + 14 + k = 0\)

For the quadratic equation \(ax^2 + bx + c = 0\) to have two equal roots, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 4\), \(b = -24\), and \(c = 14 + k\).

Substitute these values into the discriminant formula:

\((-24)^2 - 4 \times 4 \times (14 + k) = 0\)

\(576 - 16(14 + k) = 0\)

\(576 - 224 - 16k = 0\)

\(352 = 16k\)

\(k = \frac{352}{16} = 22\)

(i) To solve \(gf(x) = x\), substitute \(f(x) = 2x + 1\) into \(g(x)\):

\(gf(x) = \frac{2(2x + 1) - 1}{2x + 1 + 3} = \frac{4x + 1}{2x + 4}\)

Set \(\frac{4x + 1}{2x + 4} = x\) and solve for \(x\):

\(4x + 1 = x(2x + 4)\)

\(4x + 1 = 2x^2 + 4x\)

\(2x^2 = 1\)

\(x^2 = \frac{1}{2}\)

\(x = \frac{1}{2}\sqrt{2}\)

(ii) To find \(f^{-1}(x)\), solve \(y = 2x + 1\) for \(x\):

\(x = \frac{1}{2}(y - 1)\)

Thus, \(f^{-1}(x) = \frac{1}{2}(x - 1)\).

To find \(g^{-1}(x)\), solve \(y = \frac{2x - 1}{x + 3}\) for \(x\):

\(y(x + 3) = 2x - 1\)

\(yx + 3y = 2x - 1\)

\(yx - 2x = -3y - 1\)

\(x(y - 2) = -3y - 1\)

\(x = \frac{-3y - 1}{y - 2}\)

Thus, \(g^{-1}(x) = \frac{-1 - 3x}{x - 2}\) or \(\frac{1 + 3x}{2 - x}\).

(iii) To show \(g^{-1}(x) = x\) has no solutions, set \(\frac{-1 - 3x}{x - 2} = x\):

\(-1 - 3x = x(x - 2)\)

\(-1 - 3x = x^2 - 2x\)

\(x^2 + x + 1 = 0\)

The discriminant \(b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = -3\), which is negative, indicating no real roots.